Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 1
This print-out should have 47 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
Hence
h→0
=
10.0 points
Find an equation for the tangent to the
graph of f at the point P (2, f (2)) when
f (x) =
f ′ (2) =
f (x + h) − f (x) =
=
5
5
−
1 − 3(x + h) 1 − 3x
5(1 − 3x) − 5{1 − 3(x + h)}
(1 − 3x) {1 − 3(x + h)}
15h
.
=
(1 − 3h) (1 − 3(x + h))
Thus
15
f (x + h) − f (x)
=
.
h
(1 − 3x) (1 − 3(x + h))
3
15
= ,
2
(1 − 6)
5
so by the point slope formula an equation for
the tangent line at (2, −1) is
y+1 =
3
(x − 2)
5
which after simplification becomes
4. y = 3x − 7
3
1
5. y + x + = 0
5
5
Explanation:
If x = 2, then f (2) = −1, so we have to find
an equation for the tangent line to the graph
of
5
f (x) =
1 − 3x
at the point (2, −1). Now the Newtonian
quotient for f at a general point (x, f (x)) is
given by
f (x + h) − f (x)
.
h
First let’s compute the numerator of the Newtonian Quotient:
15
.
(1 − 3x)2
At x = 2, therefore,
5
.
1 − 3x
2
3
1. y + x + = 0
5
5
3
11
2. y = x −
correct
5
5
1
9
3. y = x −
5
5
15
(1 − 3(x + h))(1 − 3x)
f ′ (x) = lim
11
3
x−
5
5
y =
002
Find
.
10.0 points
dy
when
dx
2
3
√ + √ = 5.
y
x
1.
dy
3 x 3/2
=
dx
2 y
2.
dy
2 y 3/2
= −
correct
dx
3 x
3.
2
dy
= (xy)1/2
dx
3
4.
dy
3
= (xy)1/2
dx
2
5.
2 y 3/2
dy
=
dx
3 x
6.
3 x 3/2
dy
= −
dx
2 y
Explanation:
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 2
Differentiating implicitly with respect to x,
we see that
1 2
3 dy √ + √
−
= 0.
2 x x y y dx
Consequently,
dy
2 y 3/2
.
= −
dx
3 x
003
y = x2 − 3x
and dx/dt = 3.
dy 1.
= 5
dt x=3
dy = 9 correct
2.
dt x=3
dy = 3
3.
dt x=3
dy 4.
= 1
dt x=3
dy = 7
5.
dt x=3
Explanation:
Differentiating implicitly with respect to t
we see that
dy
dx
= (2x − 3)
= 3 (2x − 3) .
dt
dt
At x = 3, therefore,
5 cos(x)
(sin(x) + 4)2
4. f ′ (x) = −
3 sin(x) cos(x)
sin(x) + 4
5. f ′ (x) =
5 cos(x)
correct
(sin(x) + 4)2
6. f ′ (x) =
3 cos(x)
sin(x) + 4
Explanation:
By the Quotient Rule,
f ′ (x) =
(sin(x) + 4) cos(x) − (sin(x) − 1) cos(x)
.
(sin(x) + 4)2
But
(sin(x)+4) cos(x)−(sin(x)−1) cos(x) = 5 cos(x) .
Thus
f ′ (x) =
5 cos(x)
(sin(x) + 4)2
.
keywords: derivative of trig functions, derivative, quotient rule
005
10.0 points
The points P and Q on the graph of
y 2 − xy − 5y + 10 = 0
dy
= 3(3) = 9 .
dt
10.0 points
′
Determine f (x) when
f (x) =
1. f ′ (x) = −
5 sin(x) cos(x)
sin(x) + 4
3. f ′ (x) = −
10.0 points
Determine the value of dy/dt at x = 3 when
004
2. f ′ (x) =
sin(x) − 1
.
sin(x) + 4
3 cos(x)
(sin(x) + 4)2
have the same x-coordinate x = 2. Find the
point of intersection of the tangent lines to
the graph at P and Q.
5 20
1. intersect at =
correct
,
7 7
10 20
2. intersect at =
,
3 7
20
5
,−
3. intersect at =
7
7
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 3
4. intersect at =
20 5
,
7 3
5. intersect at =
5 20
,
3 7
Explanation:
The respective y-coordinates at P, Q are
the solutions of
y 2 − xy − 5y + 10 = 0
(‡)
at x = 2; i.e., the solutions of
y 2 − 7y + 10 = (y − 5)(y − 2) = 0.
Thus
and
2
10
y+ x =
3
3
respectively. These two tangent lines
5 20
,
intersect at =
.
7 7
006
10.0 points
At noon, ship A is 150 miles due west of
ship B. Ship A is sailing south at 30 mph
while ship B is sailing north at 10 mph.
At what speed is the distance between the
ships changing at 5:00 pm?
1. speed = 34 mph
P = (2, 5),
Q = (2, 2).
To determine the tangent lines we need also
the value of the derivative at P and Q. But
by implicit differentiation,
2y
dy
dy
− (x + 5)
− y = 0.
dx
dx
so
y
dy
=
.
dx
2y − x − 5
Thus
dy 5
= ,
dx P
3
2
dy = − .
dx Q
3
2. speed = 38 mph
3. speed = 32 mph correct
4. speed = 30 mph
5. speed = 36 mph
Explanation:
Let a = a(t) be the distance travelled by
ship A up to time t and b = b(t) the distance
travelled by ship B. Then, if s = s(t) is
the distance between the ships, the relative
positions and directions of movement of the
ships are shown in
By the point-slope formula, therefore, the
equation of the tangent line at P is
y−5 =
150
5
(x − 2),
3
b
s
while that at Q is
2
y − 2 = − (x − 2).
3
a
Consequently, the tangent lines at P and Q
are
y =
5
5
x+
3
3
B
A
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 4
By Pythagoras, therefore,
s2 = (a + b)2 + (150)2 .
Thus, after implicit differentiation,
da db ds
2s
,
= 2(a + b)
+
dt
dt
dt
in which case,
After implicit differentiation with respect
to x we see that
−2xy sin(x2 ) + y ′ cos(x2 ) = 0 .
Consequently,
2xy sin(x2 )
dy
=
= 2xy tan(x2 ) .
dx
cos(x2 )
008
ds
= (30 + 10)
dt
a + b
s
a+b
= 40
s
when ships A and B are travelling at respective speeds 30 mph and 10 mph.
But at 5:00 pm,
a = 150,
b = 50,
s = 250
Consequently, at 5:00 pm, the distance between the ships is increasing at a
ds = 32 mph .
speed =
dt t=5
007
10.0 points
Determine dy/dx when
y cos(x2 ) = 5 .
10.0 points
A 5 foot ladder is leaning against a wall. If
the foot of the ladder is sliding away from the
wall at a rate of 12 ft/sec, at what speed is
the top of the ladder falling when the foot of
the ladder is 4 feet away from the base of the
wall?
1. speed = 16 ft/sec correct
2. speed = 15 ft/sec
3. speed =
49
ft/sec
3
4. speed =
46
ft/sec
3
5. speed =
47
ft/sec
3
Explanation:
Let y be the height of the ladder when the
foot of the ladder is x feet from the base of
the wall as shown in figure
1.
dy
= 2xy cos(x2 )
dx
2.
dy
= 2xy tan(x2 ) correct
dx
3.
dy
= −2xy tan(x2 )
dx
4.
dy
= 2xy cot(x2 )
dx
5.
dy
= −2xy cot(x2 )
dx
x ft.
6.
dy
= −2xy sin(x2 )
dx
We have to express dy/dt in terms of x, y and
dx/dt. But by Pythagoras’ theorem,
Explanation:
5 ft.
x2 + y 2 = 25 ,
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 5
so by implicit differentiation,
2x
dx
dy
+ 2y
= 0.
dt
dt
In this case
dy
x dx
= −
.
dt
y dt
But again by Pythagoras, if x = 4, then y = 3.
Thus, if the foot of the ladder is moving away
from the wall at a speed of
dx
= 12 ft/sec ,
dt
and x = 4, then the velocity of the top of the
ladder is given by
4 dx
dy
= −
.
dt
3 dt
Consequently, the speed at which the top of
the ladder is falling is
dy speed = = 16 ft/sec .
dt
keywords: speed, ladder, related rates
009
f ′ (x) = 2x + 3
at x = 2, i.e., m = 7. On the other hand,
f (2) = 12.
Thus by the point-slope formula, an equation for the tangent line at P (2, f (2)) is
y − 12 = 7(x − 2) ,
which after simplification becomes
y = 7x − 2 .
Consequently, the tangent line at P has
y-intercept = −2
010
2
f (x) = x + 3x + 2 .
1. y-intercept = 1
2. y-intercept = −2 correct
3. y-intercept = 6
4. y-intercept = −6
5. y-intercept = 2
6. y-intercept = −1
.
10.0 points
Determine the derivative of
x
.
f (x) = 2 arcsin
3
1. f ′ (x) = √
2. f ′ (x) = √
10.0 points
Find the y-intercept of the tangent line at
the point P (2, f (2)) on the graph of
Explanation:
The slope, m, of the tangent line at the
point P (2, f (2)) on the graph of f is the
value of the derivative
3. f ′ (x) = √
4. f ′ (x) = √
5. f ′ (x) = √
6. f ′ (x) = √
6
9 − x2
3
9 − x2
2
1 − x2
6
1 − x2
2
correct
9 − x2
3
1 − x2
Explanation:
Use of
1
d
,
arcsin(x) = √
dx
1 − x2
together with the Chain Rule shows that
2
1
′
f (x) = p
.
1 − (x/3)2 3
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 6
Consequently,
f ′ (x) = √
011
2
.
9 − x2
3e3x
1. f (x) = √
correct
1 − e6x
′
3
2. f (x) = √
1 − e6x
′
3. speed = 12 cms/min
4. speed =
35
cms/min
3
5. speed =
34
cms/min correct
3
Explanation:
Let b be the length of the base and h the
height of the triangle. Then the triangle has
3e3x
3. f ′ (x) =
1 + e6x
3
1 + e6x
area = A =
1
1 + e6x
and so
1 dA
dh db
2
=
−b
dt
h
dt
dt
1
7. f (x) = √
1 − e6x
′
e3x
8. f (x) = √
1 − e6x
=
′
2 dA A dh ,
−
h dt
h dt
since b = 2A/h. Thus, when
Explanation:
Since
d
1
,
sin−1 x = √
dx
1 − x2
1
bh.
2
Thus by the Product Rule,
1 dh
db dA
b
,
=
+h
dt
2 dt
dt
e3x
5. f (x) =
1 + e6x
′
d ax
e = aeax ,
dx
the Chain Rule ensures that
3e3x
f (x) = √
.
1 − e6x
′
012
32
cms/min
3
2. speed = 11 cms/min
f (x) = sin−1 (e3x ) .
6. f ′ (x) =
1. speed =
10.0 points
Find the derivative of
4. f ′ (x) =
At what speed is the base of the triangle
changing when the height of the triangle is 3
cms and its area is 15 sq. cms?
10.0 points
The height of a triangle is increasing at a
rate of 4 cm/min while its area is increasing
at a rate of 3 sq. cms/min.
dh
= 4,
dt
and
dA
= 3,
dt
we see that
4A 2
db
3−
cms/min .
=
dt
h
h
Consequently, at the moment when
h = 3 and
A = 15 ,
the base length is changing at a
speed =
34
cms/min
3
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 7
(recall: speed is non-negative).
014
013
10.0 points
Use linear approximation
with a = 16 to
√
estimate the number 16.2 as a fraction.
1.
√
2.
√
1
16.2 ≈ 4
80
3.
√
3
16.2 ≈ 4
80
4.
√
1
16.2 ≈ 4
20
5.
√
1
16.2 ≈ 4
16
16.2 ≈ 4
1
correct
40
Find y ′ when
xy + 5x + 2x2 = 6 .
1. y ′ =
and for values of x near a
f (x) ≈ L(x) = f (a) + f ′ (a)(x − a)
4. y ′ =
5 + 2x − y
x
5. y ′ =
y + 5 + 4x
x
6. y ′ = −
Now set
f ′ (x) =
√
Then, if we can calculate
approximation
√
a+h ≈
√
1
√ .
2 x
a easily, the linear
h
a+ √
2 a
provides a very simple method via calculus
for
√ computing a good estimate of the value of
a + h for small values of h.
y + 5 + 4x
correct
x
Explanation:
Differentiating implicitly with respect to x
we see that
provides a reasonable approximation for f (x).
x,
y + 5 + 2x
x
3. y ′ = −(y + 5 + 4x)
L(x) = f (a) + f ′ (a)(x − a)
√
y + 5 + 2x
x
2. y ′ = −
Explanation:
For a general function f , its linear approximation at x = a is defined by
f (x) =
10.0 points
Thus
d
d
xy + 5x + 2x2 =
(6) .
dx
dx
(xy ′ + y) + 5 + 4x = 0 ,
and so
xy ′ = −y − 5 − 4x .
Consequently,
y′ = −
015
y + 5 + 4x
.
x
10.0 points
In the given example we can thus set
a = 16 ,
2
.
h =
10
For then
√
16.2 ≈ 4
1
.
40
Let f, g be differentiable functions. Consider the following statements:
A. If F1 (x) = {f (x) − 1/f (x)}2 , then
1
′
′
.
F1 (x) = 2f (x) f (x) +
f (x)3
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 8
Consequently,
B. If F2 (x) = f (x)g(x), then
F2′ (x) = f ′ (x)g(x) + f (x)g ′(x) .
B. True
C. If F3 (x) = f (x)/g(x), then
F3′ (x)
A. Not True
f ′ (x)g(x) + f (x)g ′(x)
=
.
g(x)2
Which of these statements are true?
C. Not True
016
10.0 points
Find the derivative of f when
√
f (x) = x(2x + 7) .
1. C only
2. B and C only
1. f ′ (x) =
4x + 7
√
x x
2. f ′ (x) =
6x + 7
√ correct
2 x
3. f ′ (x) =
6x − 7
√
2 x
4. f ′ (x) =
4x − 7
√
x x
7. none of them
5. f ′ (x) =
6x − 7
√
x x
8. A and C only
6. f ′ (x) =
4x + 7
√
2 x
3. A and B only
4. A only
5. all of them
6. B only correct
Explanation:
By the respective Product and Quotient
Rules,
d
f (x)g(x) = f ′ (x)g(x) + f (x)g ′(x)
dx
while
Explanation:
By the Product Rule
f ′ (x) =
√
2x + 7
√ + 2 x.
2 x
After simplification this becomes
d f (x)
=
dx g(x)
f ′ (x)g(x) − g ′ (x)f (x)
g(x)2
Applying these to
F (x) = f (x) −
1
f (x)
.
2
f ′ (x) =
6x + 7
2x + 7 + 4x
√
√
=
.
2 x
2 x
017
= f (x)2 − 2 +
1
,
f (x)2
Find the derivative of
f (x) =
we see that
F (x) = 2f (x) f (x) −
′
′
1
f (x)3
.
10.0 points
1. f ′ (x) = √
sin−1 (3x)
2
.
3
sin−1 (3x)
1 − 9x2
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 9
′
2. f (x) = cos (3x) sin (3x)
3. f ′ (x) = √
6
9 − x2
sin−1 (3x)
4. f ′ (x) = 6 cos (3x) sin (3x)
5. f ′ (x) = √
6. f ′ (x) = √
3
sin−1 (3x)
2
9−x
6
sin−1 (3x) correct
2
1 − 9x
Explanation:
The Chain Rule together with
a
d
sin−1 (ax) = √
dx
1 − a 2 x2
Thus the slope of the tangent line at P is
′ π
2 π
f
= 9 sec
= 18 .
4
4
By the point-slope formula, therefore, an
equation for the tangent line at P is given
by
π
,
y − 9 = 18 x −
4
which after simplification becomes
π
y = 18x + 9 1 −
.
2
019
Find f ′ (x) when
shows that
f ′ (x) = √
018
6
sin−1 (3x) .
1 − 9x2
10.0 points
Find an equation for thetangent
line to the
π π
graph of f at the point P
when
,f
4
4
f (x) =
Explanation:
π
When x = , then f (x) = 9, so P =
4
π , 9 . Now
4
′
2
f (x) = 9 sec (x) .
4x − 1
.
5x − 1
1. f ′ (x) =
5 − 4x
(5x − 1)2
2. f ′ (x) =
1
correct
(5x − 1)2
3. f ′ (x) =
20x − 4
(5x − 1)2
4. f ′ (x) =
1
5x − 1
f (x) = 9 tan(x) .
π
1. y = 10x + 2 1 −
4
π
2. y = 18x + 9 1 −
correct
2
π
3. y = 13x + 18 1 −
4
π
4. y = 14x + 3 1 −
4
π
5. y = 17x + 14 1 −
2
10.0 points
5. f ′ (x) = −
1
(5x − 1)2
Explanation:
Using the Quotient Rule for differentiation
we see that
f ′ (x) =
4(5x − 1) − 5(4x − 1)
.
(5x − 1)2
Consequently,
f ′ (x) =
020
1
.
(5x − 1)2
10.0 points
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 10
A circle of radius r has area A and circumference C are given respectively by
A = πr 2 ,
1 − 2 cos x
correct
sin2 x
2. f ′ (x) =
2 + sin x
cos2 x
3. f ′ (x) =
1 + 2 cos x
sin2 x
4. f ′ (x) =
2 − sin x
cos x
5. f ′ (x) =
1 + 2 cos x
sin x
6. f ′ (x) =
2 − sin x
cos2 x
C = 2πr .
If r varies with time t, for what value of r is
the rate of change of A with respect to t twice
the rate of change of C with respect to t?
1. r =
1. f ′ (x) =
1
2
2. r = 2π
3. r = 2 correct
4. r = 1
5. r =
Explanation:
By the Quotient Rule,
π
2
6. r = π
f ′ (x) =
Explanation:
Differentiating
(sin x) sin x − (cos x)(2 − cos x)
sin2 x
=
A = πr 2 ,
C = 2πr
(cos2 x + sin2 x) − 2 cos x
.
sin2 x
But
implicitly with respect to t we see that
dA
dr
= 2πr ,
dt
dt
dC
dr
= 2π .
dt
dt
Thus the rate of change, dA/dt, of area is
twice the rate of change, dC/dt, of circumference when
dA
dC
= 2
,
dt
dt
i.e., when
2πr
dr dr
.
= 2 2π
dt
dt
This happens when
r = 2 .
021
10.0 points
Find the derivative of
f (x) =
2 − cos x
.
sin x
cos2 x + sin2 x = 1 .
Consequently,
f ′ (x) =
1 − 2 cos x
.
sin2 x
keywords: DerivTrig, DerivTrigExam, quotient rule
022
10.0 points
Determine f ′ (x) when
x
.
f (x) = tan−1 √
3 − x2
(Hint : first simplify f .)
√
3
′
1. f (x) = √
3 + x2
x
2. f ′ (x) = √
2
x −3
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 11
1
correct
3 − x2
√
3
′
4. f (x) = √
3 − x2
x
5. f ′ (x) = 2
x +3
3. f ′ (x) = √
Explanation:
If
x
,
tan θ = √
3 − x2
then by Pythagoras’ theorem applied to the
right triangle
2. f ′ (x) = −
sec2 (x)
(8 + 3 sec(x) tan(x))2
3. f ′ (x) =
8 sec(x) + 3
(8 + 3 sec(x))2
4. f ′ (x) =
8 cos(x) + 3
(8 cos(x) + 3)2
5. f ′ (x) =
8 + 3 cos(x)
correct
(8 cos(x) + 3)2
6. f ′ (x) =
8 sin(x) + 3 cos(x)
(8 cos(x) + 3)2
Explanation:
Now,
√
3
x
θp
3 − x2
we see that
sin(x)
tan(x)
sin(x)
cos(x)
=
=
.
3
8 + 3 sec(x)
8 cos(x) + 3
8 + cos(x)
Thus
f ′ (x) =
x
sin θ = √ .
3
Thus
x f (x) = sin−1 √ .
3
Consequently,
f ′ (x) = √
=
d
1
tan−1 x =
.
dx
1 + x2
cos(x)(8 cos(x) + 3) + 8 sin2 (x)
.
(8 cos(x) + 3)2
Consequently,
1
.
3 − x2
Alternatively, we can differentiate f using the
Chain Rule and the fact that
8 sin2 (x)
cos(x)
+
8 cos(x) + 3 (8 cos(x) + 3)2
f ′ (x) =
8 + 3 cos(x)
,
(8 cos(x) + 3)2
since cos2 (x) + sin2 (x) = 1.
024
10.0 points
If y = y(x) is defined implicitly by
.
6e5y = 3xy + 6x ,
023
10.0 points
Find the derivative of f when
f (x) =
1. f ′ (x) =
tan(x)
.
8 + 3 sec(x)
sec2 (x)
8 + 3 sec(x) tan(x)
find the value of dy/dx at (x, y) = (1, 0).
1.
dy
2
= −
dx
9
2.
2
dy
=
correct
dx
9
3.
dy
5
=
dx
27
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 12
4.
dy
7
= −
dx
27
5. rate = 34 ccs/min
Explanation:
After differentiation of P V = C with respect to t using the Product Rule we see that
2
dy
= −
5.
dx
11
6.
dy
2
=
dx
11
dV
dP
d(P V )
= P
+V
= 0,
dt
dt
dt
Explanation:
Differentiating
which after rearrangement becomes
V dP
dV
= −
.
dt
P dt
6e5y = 3xy + 6x
implicitly with respect to x we see that
dy
dy
= 3y + 3x
+ 6.
30 e5y
dx
dx
In this case,
dy 5y
30e − 3x = 3y + 6 .
dx
Consequently, at (x, y) = (1, 0) we see that
2
dy
=
.
dx
9
keywords: implicit differentiation, exponential function,
025
10.0 points
Boyle’s Law states that when a sample of
gas is compressed at a constant temperature,
the pressure and volume satisfy the equation
P V = C, where C is a constant. Suppose that
at a certain instant the volume is 400 ccs,
the pressure is 80 kPa, and the pressure is
increasing at a rate of 8 kPa/min.
At what rate is the volume decreasing at
this instant?
When
V = 400 ,
P = 80 ,
therefore,
400 · 8
dV
= −
= −40 .
dt
80
Consequently, the volume is decreasing at a
rate of 40 ccs/min .
026
10.0 points
Determine f ′ (x) when
x
−1
√
f (x) = sin
.
2 + x2
(Hint : first simplify f .)
1. rate = 36 ccs/min
x
2 + x2
√
2
2. f ′ (x) =
correct
2 + x2
x
3. f ′ (x) = 2
x +2
2. rate = 40 ccs/min correct
4. f ′ (x) =
3. rate = 42 ccs/min
4. rate = 38 ccs/min
dP
= 8,
dt
1. f ′ (x) = √
1
2 + x2
√
2
′
5. f (x) = √
2 + x2
Explanation:
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 13
If
x
,
2 + x2
then by Pythagoras’ theorem applied to the
right triangle
sin θ = √
2
p 2+
θ
x
x
1
1
4. L(x) = √ − x
2 2
1
1
5. L(x) = √ + x
2 2
1 1
1− x
6. L(x) =
2
2
Explanation:
The linearization of f is the function
L(x) = f (0) + f ′ (0)x .
√
2
But for the function
we see that
x
tan θ = √ .
2
f (x) = √
the Chain Rule ensures that
Thus
x f (x) = θ = tan−1 √ .
2
1
f ′ (x) = − (2 + x)−3/2 .
2
Consequently,
Consequently,
f ′ (x) =
Alternatively, we can differentiate f using the
Chain Rule and the fact that
d
1
.
sin−1 x = √
dx
1 − x2
.
1
f ′ (0) = − √ ,
4 2
1
f (0) = √ ,
2
√
2
.
2 + x2
1
= (2 + x)−1/2 ,
2+x
and so
1 1 L(x) = √ 1 − x .
4
2
028
10.0 points
Find the value of f ′ (a) when
027
10.0 points
Find the linearization of
f (x) = √
1
2+x
f (t) =
1. f ′ (a) =
3
a+2
2. f ′ (a) =
3
correct
(a + 2)2
at x = 0.
1 1 1. L(x) = √ 1 − x correct
4
2
1 1 2. L(x) = √ 1 + x
4
2
1 1
1+ x
3. L(x) =
2
4
2t + 1
.
t+2
3. f ′ (a) = −
4
a+2
4. f ′ (a) = −
4
(a + 2)2
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 14
5. f ′ (a) = −
3
(a + 2)2
029
4
(a + 2)2
6. f ′ (a) =
Find the derivative of
f (x) = x2 sin(x) + 2x cos(x) .
Explanation:
By definition,
1. f ′ (x) = (x2 − 2) cos(x)
f (a + h) − f (a)
f (a) = lim
.
h→0
h
′
2. f ′ (x) = (x2 − 2) sin(x)
Now, for the given f ,
3. f ′ (x) = (2 + x2 ) sin(x)
2(a + h) + 1
f (a + h) =
,
a+h+2
4. f ′ (x) = (2 − x2 ) cos(x)
while
5. f ′ (x) = (x2 + 2) cos(x) correct
2a + 1
.
f (a) =
a+2
6. f ′ (x) = (2 − x2 ) sin(x)
Thus
2(a + h) + 1 2a + 1
−
f (a + h) − f (a) =
a+h+2
a+2
!
(2(a + h) + 1)(a + 2)
− (a + h + 2)(2a + 1)
(a + h + 2)(a + 2)
=
10.0 points
.
But
{2(a + h) + 1}(a + 2)
Explanation:
By the Product Rule
d 2
(x sin(x)) = 2x sin(x) + x2 cos(x) ,
dx
while
d
(2x cos(x)) = 2 cos(x) − 2x sin(x) .
dx
Consquently,
= 2h(a + 2) + (2a + 1)(a + 2) ,
f ′ (x) = (x2 + 2) cos(x) .
and
(a+h+2)(2a+1) = h(2a+1)+(a+2)(2a+1) .
keywords: DerivTrig, DerivTrigExam,
Consequently,
f (a + h) − f (a)
h
=
030 (part 1 of 2) 10.0 points
A point is moving on the graph of
h{2(a + 2) − (2a + 1)}
h(a + h + 2)(a + 2)
=
3
,
(a + h + 2)(a + 2)
in which case
f ′ (a) =
3
(a + 2)2
.
4x3 + 5y 3 = xy .
When the point is at
1 1 ,
,
P =
9 9
its x-coordinate is decreasing at a speed of 4
units per second.
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 15
What is the speed of the y-coordinate at
that time?
1. speed y-coord = −3 units/sec
031 (part 2 of 2) 10.0 points
In which direction is the y-coordinate moving at that time?
2. speed y-coord = 3 units/sec
1. direction increasing y correct
3. speed y-coord = −2 units/sec
2. direction decreasing y
4. speed y-coord = 1 units/sec
Explanation:
Since
dy
= 2,
dt
5. speed y-coord = 2 units/sec correct
at P , the y-coordinate of the point is moving
in the
Explanation:
Differentiating
4x3 + 5y 3 = xy
direction increasing y
implicitly with respect to t we see that
dx
dx
dy
dy
12x2
+ 15y 2
= y
+x .
dt
dt
dt
dt
at that time.
032
10.0 points
Thus
12x2 − y dx
dy
=
.
dt
x − 15y 2 dt
1 1
,
,
Now at P
9 9
12
1
1
2
12x − y =
=
−
(3) ,
2
(9)
9
(9)2
Find
2x3 + y 3 − 9xy − 1 = 0 .
1.
dy
2x2 − 3y
= 2
dx
y − 3x
2.
dy
2x2 + 3y
= 2
dx
y − 3x
3.
dy
2x2 − 3y
= 2
dx
y + 3x
while
x − 15y 2 =
15
1
−
9 (9)2
1
= − 2 (6) .
(9)
Hence, at P ,
1 dx
dy
= −
.
dt
2 dt
When the x-coordinate at P is decreasing at
a rate of 4 units per second, therefore,
dy
1
,
= 4
dt
2
so the
speed y-coord = 2 units/sec .
dy
when
dx
dy
3y − 2x2
4.
= 2
correct
dx
y − 3x
5.
dy
3y + 2x2
= 2
dx
y + 3x
Explanation:
We use implicit differentiation. For then
6x2 + 3y 2
dy
dy
− 9y − 9x
= 0,
dx
dx
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 16
which after solving for dy/dx and taking out
the common factor 3 gives
dy 2
2
3 (2x − 3y) +
(y − 3x) = 0 .
dx
Consequently,
3y − 2x2
dy
= 2
dx
y − 3x
10.0 points
Find the derivative of g when
g(x) = x4 cos(x) .
.
keywords: implicit differentiation, Folium of
Descartes, derivative,
033
034
10.0 points
Use linear appproximation to estimate the
value of 171/4 . (Hint: (16)1/4 = 2.)
1. g ′ (x) = x3 (4 sin(x) − x cos(x))
2. g ′ (x) = x3 (4 cos(x) − x sin(x)) correct
3. g ′ (x) = x3 (4 cos(x) + x sin(x))
4. g ′ (x) = x3 (4 sin(x) + x cos(x))
1. 171/4 ≈
31
16
5. g ′ (x) = x4 (3 cos(x) − sin(x))
2. 171/4 ≈
63
32
6. g ′ (x) = x4 (3 sin(x) − cos(x))
3. 171/4 ≈
33
16
Explanation:
By the Product rule,
4. 171/4 ≈ 2
5. 171/4 ≈
g ′ (x) = x4 (− sin(x)) + (cos(x)) · 4x3 .
65
correct
32
Consequently,
Explanation:
Set f (x) = x1/4 , so that f (16) = 2 as the
hint indicates. Then
1
df
=
.
dx
4x3/4
By differentials, therefore, we see that
f (a + ∆x) − f (a)
∆x
df ∆x =
≈
.
dx x = a
4a3/4
Thus, with a = 16 and ∆x = 1,
171/4 − 2 =
1
.
32
Consequently,
171/4 ≈
65
.
32
g ′ (x) = x3 (4 cos(x) − x sin(x)) .
035
Find
10.0 points
dy
when
dx
tan(x − y) = 3x + 2y .
1.
dy
sec2 (x − y) + 3
=
dx
sec2 (x − y) + 2
2.
2 − sec2 (x − y)
dy
=
dx
sec2 (x − y) − 3
3.
dy
2 + sec2 (x − y)
=
dx
sec2 (x − y) + 3
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 17
dy
sec2 (x − y) − 3
4.
=
dx
sec2 (x − y) − 2
5.
sec2 (x − y) − 3
dy
=
correct
dx
sec2 (x − y) + 2
6.
dy
2 − sec2 (x − y)
=
dx
sec2 (x − y) + 3
Explanation:
Differentiating implicitly with respect to x,
we see that
dy
dy 2
= 3+2 .
sec (x − y) 1 −
dx
dx
Thus by implicit differentiation,
dV
dr
= 4πr 2
= −4πr 2 ,
dt
dt
since dr/dt = −1 ins/min. When the snowball has diameter 4 inches, therefore, its radius r = 2 and
dV
= −4 (4) π .
dt
Consequently, when the snowball has diameter 4 inches, the volume of the snowball is
decreasing at a
rate = 16π cu.ins/min .
After rearranging, this becomes
dy 2
sec (x − y) + 2 = sec2 (x − y) − 3 .
dx
037
A point is moving on the graph of
6x3 + 4y 3 = xy.
Consequently,
dy
sec2 (x − y) − 3
=
.
dx
sec2 (x − y) + 2
10.0 points
When the point is at
1 1 ,
,
P =
10 10
its y-coordinate is increasing at a speed of 7
units per second.
keywords:
036
10.0 points
If the radius of a melting snowball decreases
at a rate of 1 ins/min, find the rate at which
the volume is decreasing when the snowball
has diameter 4 inches.
What is the speed of the x-coordinate at
that time and in which direction is the xcoordinate moving?
1. speed =
9
units/sec, increasing x
4
2. speed =
17
units/sec, decreasing x
8
2. rate = 18π cu.ins/min
3. speed =
9
units/sec, decreasing x
4
3. rate = 17π cu.ins/min
4. speed = 2 units/sec, increasing x
1. rate = 16π cu.ins/min correct
4. rate = 15π cu.ins/min
5. speed =
5. rate = 14π cu.ins/min
Explanation:
The volume, V , of a sphere of radius r is
given by
4
V = πr 3 .
3
7
units/sec, increasing x
4
6. speed = 2 units/sec, decreasing x
7. speed =
correct
7
units/sec,
4
decreasing x
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 18
8. speed =
17
units/sec, increasing x
8
2. speed = 15π sq. ft/sec
3. speed = 19 sq. ft/sec
Explanation:
Differentiating
4. speed = 16π sq. ft/sec correct
6x3 + 4y 3 = xy
5. speed = 16 sq. ft/sec
implicitly with respect to t we see that
18x2
dx
dy
dx
dy
+ 12y 2
= y
+x .
dt
dt
dt
dt
6. speed = 15 sq. ft/sec
7. speed = 18 sq. ft/sec
Thus
Now at P ,
x − 12y 2 dy
dx
=
.
dt
18x2 − y dt
x − 12y 2 = −
while
18x2 − y =
1
,
50
2
.
25
8. speed = 17π sq. ft/sec
Explanation:
The area, A, of a circle having radius r is
given by A = πr 2 . Differentiating implicitly
with respect to t we thus see that
dr
dA
= 2πr .
dt
dt
When
dx
1 dy
= −
.
dt
4 dt
dr
= 4,
dt
therefore, the speed at which the area of the
ripple is increasing is given by
When the y-coordinate at P is increasing at a
rate of 7 units per second, therefore,
speed = 16π sq. ft/sec
Hence, at P ,
r = 2,
dx
7
= − .
dt
4
Consequently, the x-coordinate is moving at
speed =
Find
7
units/sec ,
4
and the negative sign indicates that it is moving in the direction of decreasing x.
038
039
1. speed = 18π sq. ft/sec
dy
when
dx
tan(xy) = x − 3y .
1.
dy
1 + y sec2 (xy)
=
dx
x sec2 (xy) + 3
2.
dy
1 − y sec2 (xy)
=
correct
dx
x sec2 (xy) + 3
3.
3 + x sec2 (xy)
dy
=
dx
y sec2 (xy) − 1
4.
dy
3 − x sec2 (xy)
=
dx
y sec2 (xy) − 1
10.0 points
A rock is thrown into a still pond and causes
a circular ripple. If the radius of the ripple is
increasing at a rate of 4 ft/sec, at what speed
is the area of the ripple increasing when its
radius is 2 feet?
10.0 points
.
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 19
dy
1 − y sec2 (xy)
5.
=
dx
x sec2 (xy) − 3
6.
dy
3 − x sec2 (xy)
=
dx
y sec2 (xy) + 1
Explanation:
Differentiating implicitly with respect to x,
we see that
dy
dy = 1−3 .
sec2 (xy) y + x
dx
dx
By the Quotient and Product Rules we see
that
x+2
′
g (x) = 2
x+3
(x + 3) − (x + 2)
+ (2x − 3)
(x + 3)2
x+2
2x − 3
= 2
+
x+3
(x + 3)2
=
After rearranging, this becomes
dy x sec2 (xy) + 3 = 1 − y sec2 (xy) .
dx
2(x + 2)(x + 3) + (2x − 3)
.
(x + 3)2
But
2(x + 2)(x + 3) + (2x − 3)
= 2x2 + 12x + 9 .
Consequently,
dy
1 − y sec2 (xy)
.
=
dx
x sec2 (xy) + 3
Consequently
g ′ (x) =
keywords:
2x2 + 12x + 9
.
(x + 3)2
041
040
10.0 points
Find the derivative of
x+2
g(x) =
(2x − 3) .
x+3
10.0 points
Find dy/dx when
3x2 + 2y 2 = 5 .
1.
2x2 − 12x − 9
x+3
dy
x
=
dx
2y
2.
2. g ′ (x) =
2x2 + 12x + 9
x+3
3x
dy
=
dx
2y
3.
3. g ′ (x) =
x2 − 12x + 9
x+3
dy
3x
= −
correct
dx
2y
4.
4. g ′ (x) =
x2 + 12x − 9
(x + 3)2
3x
dy
= −
dx
y
5.
dy
= −3xy
dx
5. g ′ (x) =
2x2 + 12x + 9
correct
(x + 3)2
6.
dy
= 2xy
dx
1. g ′ (x) =
2x2 − 12x − 9
6. g (x) =
(x + 3)2
′
Explanation:
Explanation:
Diferentiating
3x2 + 2y 2 = 5
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 20
implicitly with respect to x we see that
dy
6x + 4y
= 0.
dx
Consequently,
6x
3x
dy
= −
= −
.
dx
4y
2y
042
Then by right triangle trigonometry, the angle
of inclination θ satisfies the equation
tan(θ) =
Differentiating this equation implicitly with
respect to t we see that
10.0 points
A balloon is released 3 feet away from an
observer. The balloon is rising vertically at
a rate of 3 ft/sec and at the same time the
wind is carrying it horizontally away from the
observer at a rate of 2 ft/sec. At what speed is
the angle of inclination of the observer’s line
of sight changing 3 seconds after the balloon
is released?
1. speed =
2
rads/sec
27
2. speed =
1
rads/sec
27
3. speed =
5
rads/sec
54
4. speed =
1
rads/sec correct
18
5. speed =
1
rads/sec
9
y
.
x+3
dx
dy
(x + 3) − y
dθ
dt
dt .
sec2 (θ)
=
dt
(x + 3)2
But sec2 (θ) = 1 + tan2 (θ). Thus
dy
dx
(x + 3) − y
dθ
dt
dt
=
dt
(x + 3)2 + y 2
=
3(x + 3) − 2y
rads/sec .
(x + 3)2 + y 2
Now after 3 seconds,
Explanation:
Let y = y(t) be the height of the balloon t
seconds after release and let x = x(t) be its
horizontal distance from the point of release
as shown in the figure
x(3) = 3
dx
= 6 ft/sec ,
dt
y(3) = 3
dy
= 9 ft/sec .
dt
while
Hence after 3 seconds the speed at which the
angle of inclination is changing is given by
speed =
1
rads/sec .
18
x
043
The dimensions of a cylinder are changing,
but the height is always equal to the diameter
of the base of the cylinder. If the height is
increasing at a speed of 5 inches per second,
determine the speed at which the volume, V ,
is increasing (in cubic inches per second) when
the height is 2 inches.
y
θ
3
10.0 points
x+3
1.
dV
= 14 π cub. ins./sec
dt
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 21
dV
2.
= 16 π cub. ins./sec
dt
3.
dV
= 12 π cub. ins./sec
dt
4.
dV
= 15 π cub. ins./sec correct
dt
dV
5.
= 13 π cub. ins./sec
dt
Explanation:
Since the height h of the cylinder is equal
to its diameter D, the radius of the cylinder is
1
r = h. Thus, as a function of h, the volume
2
of the cylinder is given by
V (h) = πr 2 h =
π 3
h .
4
The rate of change of the volume, therefore, is
3π
dV
2 dh
h
.
=
dt
4
dt
Now
dh
= 5 ins./sec. .
dt
y
y
x +3
dy
q
5.
=
dx
6 − xy
q
y
x +3
dy
q
6.
=
dx
6+ x
y
Explanation:
Differentiating implicitly with respect to x,
we see that
r
r
y
x dy
dy
,
+1 = 3
+
dx
x
y dx
so
dy
dx
r
r x
y
1−3
= 3
−1.
y
x
Consequently,
Find dy/dx when
045
y
x −3
dy
q
1.
=
dx
6 + xy
q
3 xy + 1
dy
q
=
2.
dx
1 − 3 xy
y
10.0 points
Find an equation for the tangent line to the
graph of
√
y + x = 6 xy .
q
q
x −1
dy
q .
=
dx
1 − 3 xy
3
dV
= 15 π cub. ins./sec .
dt
10.0 points
y
q
Consequently, when h = 2 ins.,
044
q
x −1
dy
q correct
3.
=
dx
1 − 3 xy
q
y
3
x −1
dy
q
4.
=
dx
1+3 x
3
f (x) = sec(x) + 2 cos(x)
at the point ( 13 π, f ( 31 π)).
√
√
1. 2y − 3 3x = 1 − 3π
√
√
2. 2y + 3 3x = 3π − 1
3. 2y −
√
1
3x = 3 − √ π
3
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 22
√
1
3x = 3 − √ π correct
3
√
√
5. y − 3 3x = 1 − 3π
4. y −
√
√
6. y + 3 3x = 3π − 1
f ′ (x) = sec(x) tan(x) − 2 sin(x) ,
so the slope at P is
√
3.
By the point-slope formula, therefore, an
equation for the tangent line at P is given
by
√ 1 y−3 = 3 x− π ,
3
which after rearrangement and simplification
becomes
y−
√
046
Explanation:
Since
d
1
arctan(x) =
,
dx
1 + x2
Explanation:
Since f ( 31 π) = 3, we have to find an equation for the tangent line to the graph of f at
the point P ( 13 π, 3). Now
f ′ ( 13 π) =
6. f ′ (x) = 2 sec2 (3x) tan(3x)
1
3x = 3 − √ π .
3
the Chain Rule gives
d
3
arctan(3x) =
.
dx
1 + 9x2
Using the Chain Rule yet again, therefore, we
see that
f ′ (x) =
2
arctan (3x) .
1 + 9x2
keywords: derivative, inverse tan, Chain Rule,
047
10.0 points
A street light is on top of a 12 foot pole. A
person who is 5 feet tall walks away from the
pole at a rate of 3 feet per second. At what
speed is the length of the person’s shadow
growing?
10.0 points
1. speed = 2 ft/sec
Find the derivative of
2
1
f (x) =
arctan (3x) .
3
2. speed =
15
ft/sec correct
7
3. speed =
16
ft/sec
7
2
2. f ′ (x) =
arctan (3x) correct
1 + 9x2
4. speed =
17
ft/sec
7
1
3. f ′ (x) =
arctan (3x)
9 + x2
5. speed =
18
ft/sec
7
1. f ′ (x) =
1
arctan (3x)
1 + 9x2
4. f ′ (x) =
2
arctan (3x)
9 + x2
5. f ′ (x) =
1
sec2 (3x) tan(3x)
3
Explanation:
If x denotes the length of the person’s
shadow and y denotes the distance of the
person from the pole, then shadow and the
lightpole are related in the following diagram
Version 005 – Exam Review Practice Problems NOT FOR A GRADE – alexander – (55715) 23
12
5
y
x+y
x
By similar triangles,
5
12
=
,
x
x+y
so 5y = (12 − 5)x. Thus, after implicit differentiation with respect to t,
5
dx
dy
= (12 − 5) .
dt
dt
When dy/dt = 3, therefore, the length of the
person’s shadow is growing with
speed =
15
ft/sec .
7